MATLAB gives the integer value of subtraction when the difference between numbers is large.

Hello everyone. I have a problem in my code, i will show it with an example below:

a=[1;1;1;1];

b=[10^23;10^23;10^23;10^23];

c=a-b

fprintf("%f \n",c)

Results:

c =

1.0e+23 *

-1.000000000000000

-1.000000000000000

-1.000000000000000

-1.000000000000000

-99999999999999991611392.000000

-99999999999999991611392.000000

-99999999999999991611392.000000

-99999999999999991611392.000000

When i take action with matrix it is problematic that MATLAB give the subtraction's results as integer not float, how can i change it?

Thanks for helps.

Answer 1

[Sulaymon Eshkabilov]

Specify flotaing point format accordingly, e.g.:

a=[1;1;1;1];
b=[10^23;10^23;10^23;10^23];
c=a-b
fprintf("%1999.0f \n",c)

Still it does not display the correct answer. To get a correct display you would need to use these scripts posted here: mathworks.com/matlabcentral/fileexchange/22725-variable-precision-integer-arithmetic

[Mustafa Duran]

Thanks but i will use array of "c" in another calculation in a loop, it uses false answer which is integer one.

I used vpi command but it gives that error:

If N is a double, it may be no larger than 2^53 - 1

Answer 2

[Steven Lord]

What's the distance from the elements of b to the next larger number?

b = 10^23

b = 1.0000e+23

distance = eps(b)

distance = 16777216

If you were to add 1 to b, you don't get anywhere close to the next representable floating point number.

c = b + 1;
c == b % true, and no this is NOT a bug!

ans = logical
   1

The distance between b and the next smallest representable floating point number is also much larger than 1, so if you subtract 1 you stay at b.

d = b - 1;
d == b % true and this is NOT a bug either!

ans = logical
   1

Basically what you're doing is handing Elon Musk, Jeff Bezos, or Bill Gates a $1 bill and expecting his net worth to change. It's negligible compared to what they already have. See the "Accuracy of Floating-Point Data" section on this documentation page for more information.

[Mustafa Duran]

Normally, you were right at last comment however i need this value to calculate the error term until error value is lower than 0.001 etc and after that i will break the iteration. With this rounding, MATLAB calculates error value 0 immediately, just because the not changing of values.

[Steven Lord]

This is not just MATLAB behavior. In IEEE double precision 1 is negligible compared to 10^23.

To put that in another perspective, consider something that weighs 1 kilogram, like a 1L bottle of water. Is that negligible compared to Earth's moon (with a mass of 7.3*10^22 kg according to Wikipedia)? For practical purposes, yes.

If b is just a placeholder starting value that you're using to give you an error value that's basically guaranteed to be larger than 0.001 for the first iteration, choose a smaller b.